/*
自己选择的路 ,跪着也要走完。朋友们 , 虽然这个世界日益浮躁起来,只
要能够为了当时纯粹的梦想和感动坚持努力下去 , 不管其它人怎么样,我
们也能够保持自己的本色走下去。
To the world , you will be a person , but to a person , you
will be the world .                               ——AKPower
*/
//作者:AKPower
//算法:
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <queue>
#include <cstdio>
#include <string>
#include <stack>
#include <set>
#define IOS ios::sync_with_stdio(false), cin.tie(0)
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
int dp[100010];
void init(int n)
{
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= (-1))
        {
            dp[i] += dp[i - (3 * j * j - j) / 2] * r;
            dp[i] %= MOD;
            dp[i] = (dp[i] + MOD) % MOD;
            if (i - (3 * j * j + j) / 2 >= 0)
            {
                dp[i] += dp[i - (3 * j * j + j) / 2] * r;
                dp[i] %= MOD;
                dp[i] = (dp[i] + MOD) % MOD;
            }
        }
    }
}
int solve(int n, int k)
{
    int ans = dp[n];
    for (int j = 1, r = -1; n - k * (3 * j * j - j) / 2 >= 0; j++, r *= (-1))
    {
        ans += dp[n - k * (3 * j * j - j) / 2] * r;
        ans %= MOD;
        ans = (ans + MOD) % MOD;
        if (n - k * (3 * j * j + j) / 2 >= 0)
        {
            ans += dp[n - k * (3 * j * j + j) / 2] * r;
            ans %= MOD;
            ans = (ans + MOD) % MOD;
        }
    }
    return ans;
}
int main()
{
    //IOS;
    int n;
    cin >> n;
    init(n);
    printf("%d\n", solve(n, 2));
    getchar();
    getchar();
    return 0;
}
